Comments on: Similarity – Ratios Fetish http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Li-sa http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9923 Li-sa Wed, 25 Jun 2008 06:49:29 +0000 http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9923 我的手指不肥，因为我瘦过枝火柴，只是我打字太快罢了。 我的手指不肥，因为我瘦过枝火柴，只是我打字太快罢了。

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9908 Miss Loi Tue, 24 Jun 2008 16:21:46 +0000 http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9908 For Part 4, the obvious way for many would be to do it via first finding the height of each triangle using trigonometry and then calculating their respective areas like what <b>fggffggf</b> did. However, for the benefit of those poor Sec Three students who've yet to learn trigo at this stage, here's another method utilizing the ratios obtained in Parts 2 and 3: <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/06/ratios-fetish-answer-3.gif" alt="Ratios Answer Diagram for Part 4" /> From Part 3, <m>{Area of Delta ABE}/{Area of Delta CDE} = 4/9</m> Let Δ ABE = 4 units<sup>2</sup> Let Δ CDE = 9 units<sup>2</sup> From Part 2, <m>{Area of Delta BCE}/{Area of Delta CDE} = 2/3</m> ⇒ Δ BCE = 2/3 x 9 = 6 units<sup>2</sup> Also, since Δ ABE and Δ ADE has the same base lengths (6cm and 9cm), and given they too share a common height <var>h</var> (similar to the case in Part 2), one can deduce that: <m>{Area of Delta ABE}/{Area of Delta ADE} = 2/3</m> ⇒ Δ ADE = 3/2 x 4 = 6 units<sup>2</sup> Looking at the diagram above, <m>{Area of Delta BCE}/{Area of Delta ADE} = {6 unit^2}/{6 unit^2} = 1</m> 哗……数学与 Ratios 真的好神奇啊！ For Part 4, the obvious way for many would be to do it via first finding the height of each triangle using trigonometry and then calculating their respective areas like what fggffggf did.

However, for the benefit of those poor Sec Three students who've yet to learn trigo at this stage, here's another method utilizing the ratios obtained in Parts 2 and 3:

Ratios Answer Diagram for Part 4

From Part 3, {Area of Delta ABE}/{Area of Delta CDE} = 4/9
Let Δ ABE = 4 units2
Let Δ CDE = 9 units2

From Part 2, {Area of Delta BCE}/{Area of Delta CDE} = 2/3
⇒ Δ BCE = 2/3 x 9 = 6 units2

Also, since Δ ABE and Δ ADE has the same base lengths (6cm and 9cm), and given they too share a common height h (similar to the case in Part 2), one can deduce that:

{Area of Delta ABE}/{Area of Delta ADE} = 2/3
⇒ Δ ADE = 3/2 x 4 = 6 units2

Looking at the diagram above,
{Area of Delta BCE}/{Area of Delta ADE} = {6 unit^2}/{6 unit^2} = 1

哗……数学与 Ratios 真的好神奇啊！

]]> By: fggffggf http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9791 fggffggf Fri, 20 Jun 2008 09:23:02 +0000 http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9791 Some amendment in 4: Area of △BCE = ½(6)(35-14)sin∠BEC = 63sin∠BEC cm² Area of △ADE = ½(9)(14)sin∠DEA = 63sin∠DEA cm² ∴The ratio = 63sin∠BEC : 63sin∠DEA = 1:1. Some amendment in 4:
Area of △BCE = ½(6)(35-14)sin∠BEC = 63sin∠BEC cm²
Area of △ADE = ½(9)(14)sin∠DEA = 63sin∠DEA cm²
∴The ratio = 63sin∠BEC : 63sin∠DEA = 1:1.

]]> By: fggffggf http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9734 fggffggf Wed, 18 Jun 2008 07:36:26 +0000 http://www.exampaper.com.sg/questions/e-maths/similarity-ratios-fetish#comment-9734 1. <span class="highlight">A little diagram for clarity :)</span> <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/06/ratios-fetish-answer-1.gif" alt="Answer Diagram to Part 1" /> In △ABE & △CDE, ∠ABE=∠CDE (alt ∠s, AB//CD) ∠EAB=∠ECD (alt ∠s, AB//CD) ∠BEA=∠DEC (vert opp ∠s) ∴△ABE ~ △CDE (AAA) <div class="highlight"> Great job in proving that △ABE & △CDE are <em>similar</em> using the all angles equal case (AAA - refer to this <a href="/questions/e-maths/miss-lois-unfortunate-confluence-of-factors" rel="nofollow">chart</a> for the cases of similar triangles)! Note that you don't really need to show the workings for the proof if the question didn't explicitly ask for it in your exam. However, you do need to convince yourself and state that △ABE & △CDE are <em>similar</em> first, and then obtain the ratios of the corresponding sides (see working below) in order to solve this part ;) </div> ∴BE:DE = AE:CE (corr ∠s, ~△s) 6:9 = AE:(35-AE) 2(35-AE) = 3AE AE = 14 cm. 2. <span class="highlight">Another diagram to brighten things up :)</span> <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/06/ratios-fetish-answer-2.gif" alt="Answer Diagram to Part 2" /> The altitudes of △BCE & △CDE are the same and let this altitude be h; the bases are different. ∴The ratio = (6h/2): (9h/2) = 2:3. <div class="highlight"> Yes though △BCE & △CDE are NOWHERE near similar they both share a common height <var>h</var> when calculating their respective areas - which then makes it easy to cancel <var>h</var> out when their areas are expressed as a ratio. </div> 3. ∵△ABE ~ △CDE (from 1.)∴the ratio: 62:92 = 4:9. <span class="highlight">62:92??? Fatty finger typo again???</span> <div class="highlight"> This part tests your understanding of <strong>areas of similar figures</strong> i.e. <m>A_1/A_2 = (l_1/l_2)^2</m> So since it's already proven in part 1 that △ABE & △CDE are similar, you can simply get the ratio of their areas by squaring the ratio of any two corresponding lengths of the triangles in one step BAM! <m>{Area of Delta ABE} / {Area of Delta CDE} = (6/9)^2 = 36/81 = 4/9</m> P.S. Make sure you're clear which is the numerator/denominator though! </div> 4. I can tell that this appears more difficult because students would more likely focus on the six corners given. However they may ignore the angles, in which the road to success lies. Stare at this: ∠BEC=∠DEA (vert opp ∠s) Area of △BCE = ½(6)(35-14) = 63 cm2 Area of △ADE = ½(9)(14) = 63 cm2 ∴The ratio = 63:63 = 1:1. 哗……好神奇。难怪数学这么有趣！ 1. A little diagram for clarity :)

Answer Diagram to Part 1

In △ABE & △CDE,
∠ABE=∠CDE (alt ∠s, AB//CD)
∠EAB=∠ECD (alt ∠s, AB//CD)
∠BEA=∠DEC (vert opp ∠s)
∴△ABE ~ △CDE (AAA)

Great job in proving that △ABE & △CDE are similar using the all angles equal case (AAA - refer to this chart for the cases of similar triangles)! Note that you don't really need to show the workings for the proof if the question didn't explicitly ask for it in your exam.

However, you do need to convince yourself and state that △ABE & △CDE are similar first, and then obtain the ratios of the corresponding sides (see working below) in order to solve this part ;)

∴BE:DE = AE:CE (corr ∠s, ~△s)
6:9 = AE:(35-AE)
2(35-AE) = 3AE
AE = 14 cm.

2. Another diagram to brighten things up :)

Answer Diagram to Part 2

The altitudes of △BCE & △CDE are the same and let this altitude be h; the bases are different.
∴The ratio = (6h/2): (9h/2) = 2:3.

Yes though △BCE & △CDE are NOWHERE near similar they both share a common height h when calculating their respective areas - which then makes it easy to cancel h out when their areas are expressed as a ratio.

3. ∵△ABE ~ △CDE (from 1.)∴the ratio: 62:92 = 4:9. 62:92??? Fatty finger typo again???

This part tests your understanding of areas of similar figures i.e. A_1/A_2 = (l_1/l_2)^2

So since it's already proven in part 1 that △ABE & △CDE are similar, you can simply get the ratio of their areas by squaring the ratio of any two corresponding lengths of the triangles in one step BAM!
{Area of Delta ABE} / {Area of Delta CDE} = (6/9)^2 = 36/81 = 4/9

P.S. Make sure you're clear which is the numerator/denominator though!

4. I can tell that this appears more difficult because students would more likely focus on the six corners given. However they may ignore the angles, in which the road to success lies. Stare at this: ∠BEC=∠DEA (vert opp ∠s)
Area of △BCE = ½(6)(35-14) = 63 cm2
Area of △ADE = ½(9)(14) = 63 cm2
∴The ratio = 63:63 = 1:1.
哗……好神奇。难怪数学这么有趣！

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